A logic/mathematical puzzle to start the New Year
Sunil Roy and I were strolling down the road – somewhere in Chelsea area with the rest of the family – when he posed an interesting puzzle for me. See if you can get it.
To give everybody a chance, refrain from posting the answer in the Comments section. Send me a personal message and I will put your name up if you solved it.
You have two dices. One has 1 thru 6 painted on its faces like a normal dice. The other has blank on every face. You can write down any integer number 1,2,3… on each of the blank faces of the second dice. You can even leave it blank – which would stand for 0. You are allowed to repeat the numbers (you can have two faces with the same number).
The question is: What numbers would you paint on the six faces of the second dice such that:
(*) When the two dice are rolled together, the sum total of the two faces up can be any integer between 1 and 12 AND
(*) The probability of any of those integers (meaning the sum of the two faces up being 1 or 2 or 3… or 12) is exactly the same.
You can assume that the two dices are unbiased. (It will be completely random which face will show up)
Answer to the puzzle: 0,0,0,6,6,6
The way I solved it (and there are probably many other ways) is the following:
1. The two dice, when thrown, can come up with 6×6 = 36 options.
2. For the sum to be anything between 1 and 12 and have equal chance, it means, each of the distinct sums can come up only 3 times (since there are 12 distinct sums – 1 thru 12)
3. I started with 1. For the sum to be 1 in three different possibilities, we need to have 3 zeros on the second dice.
4. Working from the other extreme, to get 12, since one die (Narayan, you oughta be proud of me) has a 6 the other one has to have 3 sixes to give 12 a chance to come up 3 different ways.
And that is the solution.
1 can come with 1 in first and 0,0,0 in second
2 can come with 2 in first and 0,0,0 in second
7 can come with 1 in first and 6,6,6 in second
8 can come with 2 in first and 6,6,6 in second